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ODE exact equations

  1. Jun 27, 2011 #1
    Can you guys point me in the right direction on the problem below?

    Solve the given initial value problem and determine at least approx. where the solution is valid:

    (2x-y)dx + (2y-x)dy= o, y(1)=3

    So I have My =-1 and Nx= -1

    x^2-xy+ h(y) => -x+h'(y) = 2y-x => h(y)= y^2

    => x^2 -xy+ y^2

    where would I go from here to solve the initial value prob?
     
  2. jcsd
  3. Jun 27, 2011 #2

    LCKurtz

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    You have the right idea but is written poorly and that is why you don't have an equation in your last step. After you have checked exactness you should arrange your work like this. You have an unknown function f(x,y) satisfying this exact differential and you know

    fx(x,y) = 2x - y
    (*) f(x,y) = x2 - xy + h(y)
    Differentiating this with respect to y and using the equation:
    -x + h'(y) = 2y - x
    h'(y) = 2y
    h(y) = y2 + C

    Substitute his in for the h(y) in (*) above which gives

    f(x,y) = x2 - xy + y2 + C

    This is the function that satisfies df(x,y) = 0 so your solution is

    x2 - xy + y2 + C = 0

    Now use your initial conditions. Notice at each step of the writeup you have an equation with an = and two sides. No sloppy use of => symbol.
     
  4. Jun 27, 2011 #3
    Thanks. I found c to be 7. But the answer in the text states y as = (x + sqrt(28-3x^2))/2. Do you have any idea how they manipulated into that?
     
  5. Jun 28, 2011 #4
    C is -7, not 7. The text is correct. Use the quadratic formula and solve for y in the equation y2 - (x)y + (x2 - 7) = 0.

    Your solution does not define a function; it is a formal solution (according to Spiegel's Applied Differential Equations) only because it satisfies the original differential equation. The way you have it now, (1, 3) and (1, -2) are both points on the curve: (1)2 - (1)(3) + (3)2 - 7 = 1 - 3 + 9 - 7 = 0 = 1 + 2 + 4 - 7 = (1)2 - (1)(-2) + (-2)2 - 7. The initial value problem states y(1) = 3 only, so take the positive branch.
     
    Last edited: Jun 28, 2011
  6. Jun 28, 2011 #5
    got it now, thanks
     
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